题解 | 数字反转

#include <iostream>
using namespace std;

int reverse(int n) {
    int t = n;
    int count = 0;
    while (t) {
        count++;
        t /= 10;
    }
    int a[count];
    t = n;
    for (int i = count - 1; i >= 0; i--) {
        a[i] = t % 10;
        t /= 10;
    }
    int res = 0;
    for (int i = count - 1; i >= 0; i--) {
        res += a[i];
        res *= 10;
    }
    return res / 10;
}

int main() {
    int n1, n2;
    int r1, r2;
    while (cin >> n1 >> n2) {
        r1 = n1 + n2;
        r2 = reverse(n1) + reverse(n2);
        if (r1 == reverse(r2)) {
            cout << r1 << endl;
        } else {
            cout << "NO" << endl;
        }
    }
}
// 64 位输出请用 printf("%lld")